The two lines L_1: vec{r}=(hat{i}+5 hat{j}+5 | vedclass

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(C) Two lines $\vec{r}=\vec{a}_1+t\vec{b}_1$ and $\vec{r}=\vec{a}_2+s\vec{b}_2$ are parallel if $\vec{b}_1=m\vec{b}_2$ for some scalar $m \in \mathbb{

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both are skew lines

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(C) Two lines $\vec{r}=\vec{a}_1+t\vec{b}_1$ and $\vec{r}=\vec{a}_2+s\vec{b}_2$ are parallel if $\vec{b}_1=m\vec{b}_2$ for some scalar $m \in \mathbb{R}$.They are perpendicular if $\vec{b}_1 \cdot \vec{b}_2 = 0$.They intersect if the shortest distance between them is $0$.Given $L_1: \vec{r}=(\hat{i}+5 \hat{j}+5 \hat{k})+t(4 \hat{i}-4 \hat{j}+5 \hat{k})$ and $L_2: \vec{r}=(2 \hat{i}+4 \hat{j}+5 \hat{k})+s(8 \hat{i}-3 \hat{j}+\hat{k})$.Here,$\vec{b}_1 = 4\hat{i}-4\hat{j}+5\hat{k}$ and $\vec{b}_2 = 8\hat{i}-3\hat{j}+\hat{k}$.Since $\vec{b}_1$ is not a scalar multiple of $\vec{b}_2$,the lines are not parallel.Check perpendicularity: $\vec{b}_1 \cdot \vec{b}_2 = (4)(8) + (-4)(-3) + (5)(1) = 32 + 12 + 5 = 49 
eq 0$. Thus,they are not perpendicular.To check if they are skew,we calculate the shortest distance $d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 	imes \vec{b}_2)|}{|\vec{b}_1 	imes \vec{b}_2|}$.$\vec{a}_2 - \vec{a}_1 = (2-1)\hat{i} + (4-5)\hat{j} + (5-5)\hat{k} = \hat{i} - \hat{j}$.$\vec{b}_1 	imes \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 4 & -4 & 5 \ 8 & -3 & 1 \end{vmatrix} = \hat{i}(-4+15) - \hat{j}(4-40) + \hat{k}(-12+32) = 11\hat{i} + 36\hat{j} + 20\hat{k}$.$(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 	imes \vec{b}_2) = (1)(11) + (-1)(36) + (0)(20) = 11 - 36 = -25 
eq 0$.Since the shortest distance is non-zero,the lines are skew lines.

The two lines $L_1: \vec{r}=(\hat{i}+5 \hat{j}+5 \hat{k})+t(4 \hat{i}-4 \hat{j}+5 \hat{k})$ and $L_2: \vec{r}=(2 \hat{i}+4 \hat{j}+5 \hat{k})+s(8 \hat{i}-3 \hat{j}+\hat{k})$ are such that

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